How Does Lens Focal Length Affect Image Size on a Wall?

[Darth Frodo]

Problem 1

A flat screen TV is place on a wall in a room. A lens of focal length 50cm is placed between the television and the opposite wall so that a sharp image with one quarter of that of the area of the television is produced on the opposite wall.

Answer:

Magnification = [itex]\frac{Image}{Object}[/itex]

M = [itex]\frac{V}{U}[/itex]

M = [itex]\frac{1}{4}[/itex]

V = [itex]\frac{U}{4}[/itex]


[itex]\frac{1}{f}[/itex] = [itex]\frac{1}{U}[/itex] + [itex]\frac{1}{V}[/itex]

[itex]\frac{1}{50}[/itex] = [itex]\frac{1}{U}[/itex] + [itex]\frac{4}{U}[/itex] (From above)

[itex]\frac{1}{50}[/itex] = [itex]\frac{5}{U}[/itex]

U = 250 cm

U = 0.25 cm


Is this right?

 

[technician]

The magnification given by m=v/u is a LINEAR magnification... this is how much lengths are magnified. If m = 2 it means lengths are 2x greater... what will that mean for AREA magnification?
Hope this helps